An equilateral triangle is inscribed in the parabola y2 = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Asked by Pragya Singh | 11 months ago |  139

1 Answer

Solution :-

Let us consider OAB be the equilateral triangle inscribed in parabola y2 = 4ax.

Let AB intersect the x – axis at point C.

Diagrammatic representation of the ellipse is as follows:

Now let OC = k

From the equation of the given parabola, we have,

So, y2 = 4ak

y = ±2\( \sqrt{ak}\)

The coordinates of points A and B are (k, \( 2 \sqrt{ak}\)), and (k, \( -2 \sqrt{ak}\))

AB = CA + CB

\( 2 \sqrt{ak}+ 2 \sqrt{ak}\)

\( 4\sqrt{ak}\)

Since, OAB is an equilateral triangle, OA2 = AB2.

Then,

k2 + (\( 2 \sqrt{ak}\))2 = (\( 4 \sqrt{ak}\))2

k2 + 4ak = 16ak

k2 = 12ak

k = 12a

Thus, AB = \( 4\sqrt{ak}\)

 = \( 4\sqrt{(a×12a)}\)

\(4\sqrt{12a^2}\)

= \( 4\sqrt{(4a×3a)}\)

= 4(2)\( \sqrt{3a}\)

\( 8\sqrt{3a}\)

Hence, the side of the equilateral triangle inscribed in parabola

y2 = 4ax is \( 8\sqrt{3a}\).

Answered by Abhisek | 11 months ago

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